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    Adrian teaches a class of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subject to this condition:
    In how many ways can he split them into three teams of four?

    Can anyone see how to do this? It's a simple problem when splitting into two teams of 6 (the precursor), but 3 teams of 4 becomes quite tough!
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    Sorry for the rough work, I was literally writing at the same time as thinking. Hopefully this solution makes sense, though I can't guarantee it's mistake free.Name:  IMG_0627.jpg
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    (Original post by TimGB)
    Sorry for the rough work, I was literally writing at the same time as thinking. Hopefully this solution makes sense, though I can't guarantee it's mistake free.
    And divide by 3!, giving 960, as the order of the teams doesn't matter.
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    (Original post by TimGB)
    Sorry for the rough work, I was literally writing at the same time as thinking. Hopefully this solution makes sense, though I can't guarantee it's mistake free.Name:  IMG_0627.jpg
Views: 99
Size:  393.0 KB
    Thank you that was really helpful - I got stuck by not realising that we had to use BOTH of the existing twin pairs in making Team 2, I guess there's a sort of pigeonhole principle-logic in that part. As the other commenter mentioned I think we need to divide by 3!, but other than that I'm pretty sure you've cracked it!
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    (Original post by MaffsIsFun)
    ...
    Since you already have a solution, as an alternative method:

    Spoiler:
    Show


    Team 1 must have representatives from 4 of the 6 pairs of twins, so
    6C4\times 2^4 possibilities.

    Team 2. Each of the "unused" teams must each be split between team2 and team3.
    So, 2C1\times 2C1 choices for those in team 2.

    And the two remaining slots contain members of the 4 teams that appeared in team1 (only one possibility for each team as the other is already assigned in team 1).
    So, 4C2 choices for them.

    Team 3 is uniquely determined from the choices made for team1 and team2.

    And since order of teams doesn't matter, divide by 3!.


    So, total no. choices =\dfrac{6C4\times 2^4\times 2C1 \times 2C1\times 4C2}{3!}=960

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    (Original post by ghostwalker)
    And divide by 3!, giving 960, as the order of the teams doesn't matter.
    Ah yes, I knew I'd forget something!
 
 
 
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